Parity Operator and Eigenvalue

Transcription

1 Parity Operator and Eigenvalue The parity operator acting on a wavefunction: PΨ(x, y, z) = Ψ(-x, -y, -z) M&S p P 2 Ψ (x, y, z) = PΨ (-x, -y, -z) = Ψ(x, y, z) P 2 = I Parity operator is unitary. If the interaction Hamiltonian (H) conserves parity [H,P] = 0 P i = P f What is the eigenvalue (P a ) of the parity operator? PΨ(x, y, z) = Ψ(-x, -y, -z) = P a Ψ(x, y, z) P 2 Ψ (x, y, z) = P a PΨ(x, y, z) = (P a ) 2 Ψ(x, y, z) = Ψ(x, y, z) P a = 1 or -1 The quantum number P a is called the intrinsic parity of a particle. If P a = 1 the particle has even parity. If P a = -1 the particle has odd parity. If the overall wavefunction of a particle (or system of particles) contains spherical harmonics we must take this into account to get the total parity of the particle (or system of particles). For a wavefunction containing spherical harmonics: PΨ(r,θ,φ) = PR(r)Y l m (θ,φ) = ( 1) l R(r)Y l m (θ,φ) The parity of the particle: P a (-1) l Parity is a multiplicative quantum number. K.K. Gan L6: Parity and Charge Conjugation 1

2 Parity of Particles The parity of a state consisting of particles a and b: (-1) L P a P b L is their relative orbital momentum. P a and P b are the intrinsic parity of the two particles. Strictly speaking parity is only defined in the system where the total momentum p = 0 since the parity operator (P) and momentum operator anticommute, Pp = -p. How do we know the parity of a particle? By convention we assign positive intrinsic parity (+) to spin 1/2 fermions: +parity: proton, neutron, electron, muon (µ - ) Anti-fermions have opposite intrinsic parity. Bosons and their anti-particles have the same intrinsic parity. What about the photon? Strictly speaking, we can not assign a parity to the photon since it is never at rest. By convention the parity of the photon is given by the radiation field involved: electric dipole transitions have + parity. magnetic dipole transitions have parity. We determine the parity of other particles (π, K...) using the above conventions and assuming parity is conserved in the strong and electromagnetic interaction. Usually we need to resort to experiment to determine the parity of a particle. K.K. Gan L6: Parity and Charge Conjugation 2

3 Parity of Pions Example: determination of the parity of the π using π - d nn. For this reaction we know many things: s π = 0, s n = 1/2, s d = 1, orbital angular momentum L d = 0, J d = 1 We know (from experiment) that the π is captured by the d in an s-wave state. The total angular momentum of the initial state is just that of the d (J = 1). The isospin of the nn system is 1 since d is an isosinglet and the π - has I = 1,-1> 1,-1> is symmetric under the interchange of particles. (see below) The final state contains two identical fermions Pauli Principle: wavefunction must be anti-symmetric under the exchange of the two neutrons. Let s use these facts to pin down the intrinsic parity of the π. Assume the total spin of the nn system = 0. The spin part of the wavefunction is anti-symmetric: 0,0> = (2) -1/2 [ 1/2,1/2> 1/2-1/2> - 1/2,-1/2> 1/2,1/2>] To get a totally anti-symmetric wavefunction L must be even (0, 2, 4 ) Cannot conserve momentum (J = 1) with these conditions! Assume the total spin of the nn system = 1. The spin part of the wavefunction is symmetric: 1,1> = 1/2,1/2> 1/2,1/2> 1,0> = (2) -1/2 [ 1/2,1/2> 1/2-1/2> + 1/2,-1/2> 1/2,1/2>] 1,-1> = 1/2,-1/2> 1/2,-1/2> To get a totally anti-symmetric wavefunction L must be odd (1, 3, 5 ) L = 1 consistent with angular momentum conservation: nn has s = 1, L = 1, J = 1 3 P 1 Parity of the final state: P n P n (-1) L = (+)(+)(-1) 1 = - Parity of the initial state: P π P d (-1) L = P π (+)(-1) 0 = P π Parity conservation: P π = - K.K. Gan L6: Parity and Charge Conjugation 3

4 Spin Parity of Particles There is other experimental evidence that the parity of the π is -: The reaction π - d nnπ 0 is not observed. The polarization of γ s from π 0 γγ. Spin-parity of some commonly known particles: State Spin Parity Particle pseudoscalar (0 - ) 0 - π, K scalar (0 + ) 0 + a 0, Higgs (not observed yet) vector (1 - ) 1 - γ, ρ, ω, φ, ψ, Υ pseudovector (axial vector) (1 + ) 1 + a 1 K.K. Gan L6: Parity and Charge Conjugation 4

5 θ τ Puzzle How well is parity conserved? Very well in strong and electromagnetic interactions (10-13 ) Not at all in the weak interaction! In the mid-1950 s it was noticed that there were 2 charged particles that had (experimentally) consistent masses, lifetimes, and spin = 0, but very different weak decay modes: θ + π + π 0 τ + π + π - π + The parity of θ + = + while the parity of τ + M&S p = -. Some physicists said the θ + and τ + were different particles, and parity was conserved. Lee and Yang said they were the same particle but parity was not conserved in weak interaction! Awarded Nobel Prize when parity violation was discovered. K.K. Gan L6: Parity and Charge Conjugation 5

6 Parity Violation in β-decay Classic experiment of Wu et. al. (Phys. Rev. V105, Jan. 15, 1957) looked at β spectrum: Co Ni * + e +ν e Ni * Ni * +γ(1.17)+γ(1.33) Parity transformation reverses all particle momenta while leaving spin angular momentum (rxp) unchanged. Parity invariance requires equal rate for (a) and (b). Fewer electrons are emitted in the forward hemisphere. An forward-backward asymmetry in the decay. 3 other papers reporting parity violation published within a month of Wu et. al.! 0.01 K p v p e- β counting rate + + depends on <J> p e which is under a p e- p v parity transformation 60 Co 60 Ni * 60 Ni * B-field J = 5 J = 4 J z = 1 J = 4 J z = 1 K.K. Gan L6: Parity and Charge Conjugation 6

7 Charge Conjugation (C) turns particles into anti-particles and visa versa. C(proton) anti-proton C(anti-proton) proton C(electron) positron C(positron) electron The operation of Charge Conjugation changes the sign of all intrinsic additive quantum numbers: electric charge, baryon #, lepton #, strangeness, etc. Variables such as momentum and spin do not change sign under C. The eigenvalues of C are ± 1: C ψ = ψ = C ψ ψ C 2 ψ = C ψ 2 ψ = ψ C ψ 2 =1 Charge Conjugation C ψ is sometimes called the charge parity of the particle. Like parity, C ψ is a multiplicative quantum number. If an interaction conserves C C commutes with the Hamiltonian: [H,C] ψ> = 0 Strong and electromagnetic interactions conserve C. Weak interaction violates C conservation. M&S p K.K. Gan L6: Parity and Charge Conjugation 7

8 Charge Conjugation of Particles Most particles are not eigenstates of C. Consider a proton with electric charge q. Let Q be the charge operator: Q q> = q q> CQ q> = qc q> = q -q> QC q> = Q -q> = -q -q> [C,Q] q> = [CQ QC] q> = 2q -q> C and Q do not commute unless q = 0. We get the same result for all additive quantum numbers! Only particles that have all additive quantum numbers = 0 are eigenstates of C. e.g. γ, ρ, ω, φ, ψ, Υ These particle are said to be self conjugate. K.K. Gan L6: Parity and Charge Conjugation 8

9 Charge Conjugation of Photon and π o How do we assign C ψ to particles that are eigenstates of C? Photon: Consider the interaction of the photon with the electric field. As we previously saw the interaction Lagrangian of a photon is: L EM = J u A u J u is the electromagnetic current density and A u the vector potential. By definition, C changes the sign of the EM field. In QM, an operator transforms as: CJC -1 = -J Since C is conserved by the EM interaction: CL EM C -1 = L EM CJ u A u C -1 = J u A u CJ u C -1 CA u C -1 = -J u CA u C -1 = J u A u CA u C -1 = -A u The photon (as described by A) has C = -1. A state that is a collection of n photons has C = (-1) n. π o : Experimentally we find that the π o decays to 2 γs and not 3 γs. BR(π 0 γγγ) BR(π 0 γγ) < This is an electromagnetic decay so C is conserved: C π = (-1) 2 = +1 Particles with the same quantum numbers as the photon (γ, ρ, ω, φ, ψ, Υ) have C = -1. Particles with the same quantum numbers as the π o (η, η ) have C = +1. K.K. Gan L6: Parity and Charge Conjugation 9

10 Handedness of Neutrinos Assuming massless neutrinos, we find experimentally: All neutrinos are left handed. All anti-neutrinos are right handed. Left handed: spin and z component of momentum are anti-parallel. Right handed: spin and z component of momentum are parallel. This left/right handedness is illustrated in π + l + ν l decay: BR(π + e + ν e ) BR(π + µ + ν µ ) = If neutrinos were not left handed, the ratio would be > 1! S π W S + e,µ ν p p π = 0 e,µ p ν S π = 0 M&S p v e + Angular momentum conservation forces the charged lepton (e, µ) to be in wrong handed state: a left handed positron (e + ). The probability to be in the wrong handed state ~ m 2 l BR(π + e + ν e ) BR(π + µ + ν µ ) = m 2 e m π m e 2 m µ m 2 = (1.230± 0.004) π m µ Handedness ~ 2x10-5 Phase space ~ 5 K.K. Gan L6: Parity and Charge Conjugation 10

11 In the strong and EM interaction C and P are conserved separately. In the weak interaction we know that C and P are not conserved separately. The combination of CP should be conserved! Consider how a neutrino (and anti-neutrino) transforms under C, P, and CP. Experimentally we find that all neutrinos are left handed and anti-neutrinos are right handed. σ p ν Charge Conjugation and Parity P right handed ν σ p C CP C σ p left handed ν P ν σ p CP should be a good symmetry. K.K. Gan L6: Parity and Charge Conjugation 11

12 Neutral Kaons and CP violation In 1964 it was discovered that the decay of neutral kaons sometimes (10-3 ) violated CP! The weak interaction does not always conserve CP! In 2001 CP violation was observed in the decay of B-mesons. M&S p CP violation is one of the most interesting topics in physics: The laws of physics are different for particles and anti-particles! What causes CP violation? It is included into the Standard Model by Kobayashi and Maskawa (Nobel Prize 2008). Is the CP violation observed with B s and K s the same as the cosmological CP violation? To understand how CP violation is observed with K s and B s need to discuss mixing. Mixing is a QM process where a particle can turn into its anti-particle! As an example, lets examine neutral kaon mixing first (B-meson mixing later): K 0 = s d K 0 = sd In terms of quark content these are particle and anti-particle. The K 0 has the following additive quantum numbers: strangeness = +1 charge = baryon # = lepton # = charm = bottom = top = 0 I 3 = -1/2 The K 0 s isospin partner is the K + (I 3 changes sign for anti-particles.) The K 0 and K 0 are produced by the strong interaction and have definite strangeness. They cannot decay via the strong or electromagnetic interaction. K.K. Gan L6: Parity and Charge Conjugation 12

13 Neutral Kaons, Mixing, and CP violation The neutral kaon decays via the weak interaction, which does not conserve strangeness. Let s assume that the weak interaction conserves CP. The K 0 and K 0 are not the particles that decay weakly since they are not CP eigenstates: P K 0 = K 0 and P K 0 = K 0 C K 0 = K 0 and C K 0 = K 0 CP K 0 = K 0 and CP K 0 = K 0 We can make CP eigenstates out of a linear combination of K 0 and K 0 : ( ) ( ) K 1 = 1 2 K 0 + K 0 K 2 = 1 2 K 0 K 0 This is just a QM system in two different basis. CP K 1 = K 1 and CP K 2 = K 2 If CP is conserved in the decay of K 1 and K 2 then we expect the following decay modes: K 1 two pions (π + π - or π 0 π 0 ) (CP = +1 states) K 2 three pions (π + π - π 0 or π 0 π 0 π 0 ) (CP = -1 states) M&S p In 1964 it was found that every once in a while ( 1/500) K 2 two pions! K.K. Gan L6: Parity and Charge Conjugation 13

14 Neutral Kaons The K 0 and K 0 are eigenstates of the strong interaction. These states have definite strangeness, are not CP eigenstates They are particle/anti-particle. They are produced in strong interactions (collisions) e.g. π p K 0 Λ or K 0 Σ 0 The K 1 and K 2 are eigenstates of the weak interaction, assuming CP is conserved. These states have definite CP but are not strangeness eigenstates. Each is its own anti-particle. These states decay via the weak interaction and have different masses and lifetimes. K 1 π 0 π 0 K 2 π 0 π 0 π : Gell-Mann/Pais pointed out there was more decay energy (phase space) available for K 1 than K 2. K 1 and K 2 lifetimes should be very different. m K - 2m π 219 MeV/c 2 m K - 3m π 80 MeV/c 2 m 2 - m 1 = 3.5x10-6 ev Expect K 1 to have the shorter lifetime. The lifetimes were measured to be: τ 1 9x10-11 sec ( ) τ 2 5x10-8 sec (Lande et. al., Phys. Rev. V103, 1901 (1956)) We can use the lifetime difference to produce a beam of K 2 s from a beam of K 0 s. Produce K 0 s using π - p K 0 Λ. Let the beam of K 0 s propagate in vacuum until the K 1 component dies out. 1 GeV/c K 1 travels on average 5.4 cm 1 GeV/c K 2 travels on average 3100 cm Need to put detector m away from target. K.K. Gan L6: Parity and Charge Conjugation 14

15 How do we look for CP violation with a K 2 beam? Look for decays that have CP = +1: K 2 π + π - or π 0 π 0 Experimentally we find: K 2 π + π - ~ 0.2% of the time K 2 π 0 π 0 Christenson et. al. PRL V13, 138 (1964) ~ 0.1% of the time Can also look for differences in decays that involve matter and anti-matter: δ(e) = BR(K 2 π e + ν e ) BR(K 2 π + e ν e ) BR(K 2 π e + ν e )+ BR(K 2 π + e = 0.333± ν e ) Mixing and CP violation δ(µ) = BR(K 2 π µ + ν µ ) BR(K 2 π + µ ν µ ) BR(K 2 π µ + ν µ )+ BR(K 2 π + µ = 0.303± ν µ ) Nature differentiates between matter and antimatter! CP violation has recently (2001) been unambiguously measured in the decay of B-mesons. This is one of the most interesting areas in HEP (will be discussed later). Observing CP violation with B-mesons is much more difficult than with kaons! B S and B L have essentially the same lifetime No way to get a beam of B L and look for forbidden decay modes. It is much harder to produce large quantities of B-mesons than kaons. K.K. Gan L6: Parity and Charge Conjugation 15